Problem 04 | Evaluation of Integrals

Problem 04
Evaluate   $\displaystyle \int_0^\infty \dfrac{e^{-t}\sin t}{t} ~ dt$.

Solution 04
From Problem 01 | Division by t:
$\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$

By first shifting property:

Problem 03 | Evaluation of Integrals

Problem 03
Find the value of   $\displaystyle \int_0^\infty te^{-3t} \sin t ~ dt$

Problem 02 | Evaluation of Integrals

Problem 02
Find the value of   $\displaystyle \int_0^\infty \dfrac{\sin t ~dt}{t}$.

Problem 01 | Evaluation of Integrals

Problem 01
Evaluate   $\displaystyle \int_0^\infty \dfrac{e^{-3t} - e^{-6t}}{t} ~ dt$

Evaluation of Integrals

If   $F(s) = \mathcal{L}\left\{ f(t) \right\}$,   then   $\displaystyle \int_0^\infty e^{-st} f(t) \, dt = F(s)$.

Taking the limit as   $s \to 0$,   then   $\displaystyle \int_0^\infty f(t) \, dt = F(0)$   assuming the integral to be convergent.

Laplace Transform of Intergrals

If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$


Laplace Transform of Derivatives

For first-order derivative:
$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$

For second-order derivative:
$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$

For third-order derivative:
$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$

For nth order derivative:

$\mathcal{L} \left\{ f^n(t) \right\} = s^n \mathcal{L} \left\{ f(t) \right\} - s^{n - 1} f(0) - s^{n - 2} \, f'(0) - \dots - f^{n - 1}(0)$


Division by t | Laplace Transform

Division by $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$


provided   $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$   exists.

Change of Scale Property | Laplace Transform

Change of Scale Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,

$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$


Second Shifting Property | Laplace Transform

Second Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   and   $g(t)
= \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$


$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$



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